(2) Electric Flux and Gauss’s Law

Electric flux and Gauss’s law

1 Institute of Life Long Learning, University of Delhi, Delhi

Discipline Course-I

Semester-II

Paper No: Electricity and Magnetism

Lesson: Electric flux and Gauss’s law

Lesson Developer: Mr. Jasmeet Singh

College/ Department: Keshav Mahavidyalaya,

University of Delhi



Table of Contents

Chapter 2. Electric flux

2.1. Electric field lines

2.2. Properties assigned to electric field lines

2.3. Electric flux

2.4. Gauss’s law

2.5. Proof of Gauss’s theorem using the concept of solid angle

2.6. Differential form of Gauss’s law

Questions

References



Learning Objectives

After going through this chapter, the student would be able to

Draw the electric field lines for some simple charge distribution

Know the general meaning of the flux ‘ of a vector field in terms of the concept of

‘rate of flow’

Define the term ‘Electric flux’

State Gauss law and prove it.

Know the importance and significance of the concept of ‘Gaussian surface’ in using Gauss law in practical situations.

2.1 Electric Field Lines

“A field line is an imaginary line drawn in such a way that the direction of tangent drawn

at any point is the same as the direction of the field at that point.” As shown in figure

2.1.

Figure 2.1 Direction of electric field vectors.

2.2 Properties Assigned to Electric Field Lines:

1. The magnitude of the field is indicated by the density of the field lines i.e. it is

strong near the center where the field lines are close together and weak farther

out, where they are relatively far-apart (fig. 2.2).

A C

B

Field vectors



Figure: 2.2 Electric field due to a point charge.

2. Field lines begin on positive charges and end on negative ones (fig. 2.3).

Figure: 2.3 Electric field between two opposite charges.

+Q

-Q

+Q



Figure: 2.4 Electric field between two similar charges.

3. Field lines can never cross, because at the point of intersection the field would

then need to have two different directions at once. This is not permissible (fig.

2.5).

Figure: 2.5 Two different directions of electric field vectors at the crossing, which is not

permissible.

+Q

+Q



Value addition: Did you Know

Gauss’s Divergence Theorem

Body text:

Gauss’s Divergence Theorem

The surface integral of the normal component of a vector field F taken over a

closed surface S enclosing a volume V is equal to the volume integral of the

divergence of the vector F taken through the volume V.

F.dSS

òò = (Ñ.V

òòò F)dV

Reference: David J. Griffiths, Introduction to Electrodynamics, 3rd

edition.

Gauss’s law has been discussed in detail in separate chapter specially devoted to this law.

2.3 Electric flux

To understand the concept of ‘flux’, imagine a vector function which represents the

velocity of motion of a fluid (say, water flowing in a river), where the velocity varies

from one place to another but is constant in time at any one position. Denote this vector

field by v , measured in m/s. Then, if a is the oriented area in square meters of a frame

lowered into the water, v.a is the rate of flow of water through the frame in cubic

meters per second.

Figure: 2.6 Motion of a fluid (say, water flowing in a river): The flux through an area a

is v.a , where v is the velocity of the fluid.



In figure 2.6, the flux through an area a is v.a , where v is the velocity of the fluid. The

flux is the volume of fluid passing through the area, per unit time. Through this example

of flow of a liquid as the rate of flow of a vector field, we can generalize the defination of

‘flux’ for any vector field.

In general, for any vector field the flux is defind as the rate of flow of the vector field

through a given area. In electrostatics, consider the simplest case, the field due to an

isolated positive point charge q and let the surface be a sphere of radius r centered on

the point charge. The magnitude of E at every point on the surface is constant

20

1 QE

4π r

and its direction is the same as that of the outward normal at that point.

So the flux ϕ through this surface

ϕ = E (Total area) = 2

0

1 Q

4π r

(4πr2) =

0

Q

The flux is independent of the size of the sphere.

The Electric flux can be regarded as proportional to the number of field lines passing

through unit area. It is measure of the number of field lines passing through S or the

density of field lines (number of field lines per unit area). Thus it can be viewed as a

measure of the total charge enclosed in the given surface .

Figure 2.7 Electric flux: Electric field lines through passing through area S.

The flux of Ethrough surface S (fig. 2.7)

FE = (E.n

^

)

S

ò da

(Eq. 2.1)

where da is infinitesimal area and

^n is the unit area vector which is perpendicular to da.




What are Spherical Polar coordinates?

Body text:

Spherical Polar coordinates (r,θ,ϕ)

r is the distance from the origin.

θ is the polar angle (the angle down from the z-axis)

ϕ is the azimuthal angle (the angle around from the x-axis)

Animation: ani4.swf

In the figure ^ ^ ^

, ,r constitute an orthogonal basis set (just like ^ ^ ^

x , y, z ) and any

vector A can be expressed in terms of them. The direction of the unit vectors ^ ^ ^

, ,r are along their increasing direction.

A = Ar r^

+ Aqq^

+ Af f^

,rA A and A are the radial, polar and azimuthal components of A .

In terms of Cartesian unit vectors



^ ^ ^ ^

^ ^ ^ ^

^ ^ ^

sin cos i sin sin j cos k

cos cos i cos sin j sin k

sin i cos j

r

An infinitesimal displacement in the ^

r direction is simply dr

dlr=dr


direction is

dlθ=r.dθ


direction is

dlϕ=r.sinθ.dϕ

Thus the infinitesimal displacement

dl = dr r^

+ r.dqq^

+ r.sinq.dff^

dV=dlr.dlq.dlf

Animation: ani1.swf Animation: ani2.swf



Animation: ani3.swf

The infinitesimal volume element dV, in spherical coordinates, is the product of the

three infintesimal displacements:

dV=dl .dl .dlr

2.sin . . .dr d dr

If we are integrating over the surface of a sphere, then r is constant, where θ and ϕ change, so

1

^ ^2dA dl .dl .sin . .r d d rr

When the surface lies in the xy plane, θ is constant while r and ϕ vary, so

2

^ ^

dA dl .dl . .r r dr d

Here r ranges from 0 to ∞

θ ranges from 0 to π

ϕ ranges from 0 to 2π

2.4 Gauss’s law

Here we consider that the field of an isolated positive point charge q is represented by

lines radiating out in all directions. Suppose we imagine this charge as surrounded by a

spherical surface of radius R, with the charge at its center. The area of this imaginary

surface is 4πR2, so if the total number of field lines, emanating from q, is N, then the

number of lines per unit surface area on the spherical surface is N/4πR2. We imagine a

second sphere concentric with the first, but with radius 2R. Its area is 4π(2R)2 =16πR2,



and the number of lines per unit area on this sphere is N/16πR2, one-fourth the density

of lines on the first sphere. This correspondes to the fact that, at distance 2R, the field

has only one-fourth the magnitude it has at distance R, and verfies the statement that

the density of lines is proportional to the magnitude of the field.

It is important to note that the total number of lines at distance 2R is the same as that

at a distance R. This is because whereas the field is inversely proportional to R2, the area

of the sphere is proportional to R2. Hence the product of the two is independent of R. In

mathematical terms,

02

1 qE

4π R

and the surface area is

A=4πR2

Hence the product of the two is

0

qEA

This is independent of R and depends only on the charge q.

Let us surround the sphere of radius R by a surface of irregular shape as shown in figure

2.8. Consider a small areal element ∆A; we note that this area is larger than the

corresponding element on a spherical surface at the same distance from q. If the normal

to the surface makes an angle θ with the radial line from q, the two sides of the area,

projected on the spherical surface, are foreshortened by a factor cosθ. Thus the

quantity, corresponding to E.∆A for the spherical surface, is E.∆Acosθ for the irregular

surface.

Figure 2.8 Electric field components on an irregular shape surface surrounding a point

charge q.

Now we may divide the entire surface into small elements ∆A, compute the quantity

E.∆Acosθ for each, and sum the results which will be again equal to q/∈0 for the

irregular surface. This follows from the result obtained above



0

qE. Acosθ =

The result holds no matter whatever the shape of the surface may be. It is necessary,

however, that the surface must be a closed surface enclosing the charge q.

In the limit when the area elements become very small, ∆A→0. In this limit, the sum

becomes an integral, called the surface integral of Ecosθ

0

S

qEcosθ.dA

(Eq. 2.2)

The circle on the integral sign reminds us that the integral is always taken over a closed

surface enclosing the charge q. Since E.cosθ is the component of E perpendicular to the

surface at each point , we may write E Ecosθ

0

S

1E dA q

(Eq. 2.3)

The quantity E dA EcosθdA also equals the electric flux d through the area dA.

Hence the total flux

d E ^ dA =ò E cosq.dA = E.dAòò0

q=

(Eq. 2.4)

Thus total flux, of the electric field, out of a closed surface is proportional to the charge

enclosed.

0

q=

Now it is easy to generalize the above results to any charge distribution. The total

electric field E, at a point on the surface, is the vector sum of the fields produced by the

individual charges and, therefore, the quantity Ecosθ.dA is the sum of the contributions

from these charges:

0S

1E dA q

The total charge enclosed Q= q

S

E.dA

0

Q

(Eq. 2.5)

Thus the Gauss’s law states that “The surface integral of the normal component of the

electric field ( E) over any closed surface (S), is equal to 1/∈0 times the total charge (Q)

contained in the volume of space bounded by that closed surface.”

Here we need to write about the need for s ‘Gaussian Surface’. Gauss’s law is applicable

for any closed surface but the Gaussian surface has to be a special closed surface for a

given charge distribution.



In evaluating the surface integral of E over a closed surface, we may divide the

surface into a number of elements. The integral, over the entire surface, is the sum of

the integrals over each element. There are many cases of practical importance where

symmetry considerations simplify the evaluation of the integral. The following

observations are also useful:

1. If E is at right angles to a surface of area A at all points and has the same

magnitude at all points of the surface, then E = E =constant and E ^ dA = EAò

2. If E is parallel to a surface at all points, E =0 and the integral is zero.

3. If E=0 at all points of a surface, the integral is zero.

4. The ‘Gaussian surface’ to which Gauss’s law is applied need not be a real physical

surface , such as the surface of a soild body. In most applications of this law, we

often use (i) an imaginary surface that may be in empty space, (ii) a (suitable)

surface embedded in a solid body, or (iii) a (suitable) surface that may be partly

in space and partly within a body.


Difference between an angle and solid angle.

Body text:

Angle: Arc AB

θ =radius r

The unit angle is the angle subtended by an arc of length equal to its radius. It is

known as one radian.

Solid angle:

2

2

dSArcdw= =

radius r

The unit solid angle is that angle subtended at the center of a sphere by an area r2

and is known as steradian. The total surface area of the sphere is 4πr2; hence the

total solid angle subtended at its center is 4π steradians.

θ

A B

O

r



Solid angle is the analogue, in three dimensions, of the usual angle in two

dimensions.

For a surface S of irregular shape, dS and dS’ subtend the same solid angle. Thus

the solid angle, subtended by any surface dS, at a point O, a distance r away, is

given by

2

dS.cosθd =

rΩ

where dS’=dScosθ is the projection of the surface dS perpendicular to the radius

vector r from the point O.

Reference: Electricity and Magnetism By D C Tayal (Himalaya Publishing House,1988).

2.5 Proof of Gauss’s theorem using the concept of Solid Angle

To prove Gauss’s theorem we begin with the case when the source, producing the field,

is a point charge q present in a volume enclosed by a surface S of irregular shape. We

consider an infinitesimal element dS of the surface S as shown in figure 2.9. The flux,

through dS, is given by:

d EcosθdS (Eq. 2.6)

where field 0

2

1 qE

4π r

dΩ

dw

dS

O



0 02 2

1 q q dScosθ d cosθdS= dω

4π 4πr r

Figure 2.9 Electric flux through an elementary area dS of a closed surface surrounding a

point charge.

Here 2

dScosθ

rd Ω is the solid angle subtended by the surface dS at the point where the

point charge is located. Thus the total flux, through the entire closed surface, S

0 0S S S

q qEcosθdS d d

4π 4π

Ω Ω

where

S

d Ω is the total solid angle subtended by the surface S. We know that the total

solid angle, subtended by a closed surface, around a point in space, is 4 π .

E.dSò =q

4π Î 0

´ 4π=0

q

(Eq. 2.7)

This is the Gauss’s theorem for a single point charge present inside the surface S.

Consider now a charge q lying outside the closed surface. We construct a cone of solid

angle dω with q as vertex (fig. 2.10). Let dS1 and dS2 be the areas inetrcepted by the

cone at A1 and A2. The projection of these surfaces, perpendicular to the radius vector,

are dS1cosα1 and dS2cosα2 and

1 2

1 2

1 2

2 2

ds cosα ds cosαd

r r Ω



Figure 2.10 Electric flux through a closed surface due to a point charge present outside.

The flux through dS1 is directed inwards, as the angle (π- α1), between the directions of

E and 1

^

n , is obtuse. The outward flux is

-E1 dS1cosα1+E2 dS2cosα2=1 2

0 01 2

1 2

2 2

-q ds cosα q ds cosα0

4π 4πr r

Thus the net flux through the entire surface due to charge outside is zero.

2.6 Differential form of Gauss’s law

Gauss’s law, in integral form, is expressed as

enclosed

0S

QE.da

(Eq. 2.8)

Applying divergence theorem to the L.H.S. of above equation, we get

S V

E.da .E dV

Also enclosedQ can be expressed in terms of the charge density ρ :

enclosed

V

Q = ρ dV

Thus Gauss’s law becomes

0V V

ρ.E dV dV

For any arbitrary volume, the integrands must be equal. Hence

Ñ.E =r

Î 0 (Eq. 2.9)

This is Gauss’s law in differential form. We may state this as follows:

The divergence, of the electric field, at any point, is 01 times the volume charge

density at that point.

Question Number Type of question

1 Subjective

Question



(a) A point charge Q is located at the center of a cube of edge length d. What is

the value of

S

E.da

over one face of the cube?

(b) The charge Q is moved to one corner of the cube. What is now the value of

the flux of E

through each of the nearby faces of the cube?

Solution

(a) The magnitude of E at every point on the surface is constant

20

1 QE

4π r

. So the total flux ϕ through all the six faces

ϕ = E (Total area)=

S

E.da

= 0

Q

From the defination, the electric flux is proportional to the number of

field lines passing through unit area. So the flux through a single face

is

ϕ1 = 0

Q

6

(b) Now the point charge is present at one corner of the cube. Hence the

flux contribution to the cube will be 1/8th of total flux (Q/∈0).

The effect of field lines, due to a point charge, drops with distance.

Hence the flux, due to a point charge, present at one corner of the

cube, will pass only through three faces near to the point charge.

Hence the flux through a single face is

ϕ1 = 0 0

1 1 Q Q. .

8 3 24

.


2 Subjective

Question

Suppose the electric field, in some region, is found to be E

= kr3^

r , in spherical

coordinates (k is some constant)

(a) Find the charge density ρ.

(b) Find the total charge contained in a sphere of radius R, centered at the



origin.

Solution

(a) Using differential form of Gauss’s law

Ñ.E =r

Î 0

k3r2 = 0

ρ

02ρ = 3k r

(b) We have a relation between total chrage and chrage density, given

by

V

Q = ρ dV

Using sperical coordinates

0

R π 2π

r 0 θ 0 0

Q = 3k

r2.r2 sinθ dr dθ dϕ

0

12Q = kπ

5 R5


3 Subjective

What is the total outward flux from the cube (0<x,y,z<1 meter), containing a

volume charge distribution ρ = 16xyz μC/m3.

Solution According to Gauss’s law,

Total outward flux= Total charge enclosed/∈0 = Q/∈0 =

V

ρdV /∈0



ϕ = 0

16

× 10-6

1 1 1

0 0 0

xyz dx dy dz

ϕ = 0

2

References:

1. David J. Griffiths, Introduction to Electrodynamics, 3rd edition.

2. Electricity and Magnetism By D C Tayal (Himalaya Publishing House,1988)

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(2) Electric Flux and Gauss’s Law